[Ni{(CO)}_4] is [Pb. CET 2000]
A) $ sp^{3} $
B) $ d^{2}sp^{3} $
C) $ sp^{3}d $
D) $ sp^{2} $
Correct Answer: A
In the compound $ [Ni{{(CO)}_4}] $ Oxidation number of $ Ni=0 $ Co-ordination number of $ Ni=4 $ (CO (carbonyl) is a strong ligand so it causes pairing of electrons)
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